First off I'd like to say that when you use any of these three methods you're finding values of x where the function crosses the x-axis.

An example of factorization would be this:
f(x)= x^2+4x+4

If we set this equal to 0 (because this means the function is on the x-axis) we get:
0 = x^2+4x+4

To factor this we are just looking for a way to essentially make it easier to find the zero's or x-intercepts.

This particular equation factors into (x+2)(x+2) so we have
0 = x^2+4x+4 = (x+2)(x+2)

So the value of x where this quadratic will cross the x-axis happens when x= -2

Because then we get 0 = (-2+2)(-2+2)

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If we use the quadratic formula for the same problem f(x)=x^2+4x+4 we will indeed get the same answer.

The quadratic formula itself goes like this:
For some function f(x) = ax^2+bx+c

[tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

So we can go back to our problem f(x) = x^2+4x+4 and we get

[tex] x = \frac{-4 \pm \sqrt{4^2-4(1)(4)}}{2(1)}[/tex]

[tex] x = \frac{-4 \pm \sqrt{0}}{2}[/tex]

[tex] x = \frac{-4}{2} = -2[/tex] Which is exactly what we got above by factoring.

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To find the zeros graphing requires a graphing calculator.

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Now see if you can solve those three equations you posted. I'll give you a hint: the first can be factored and the quadratic formula would be really easy to use on the second problem, and the third you can graph.

You already learned about factoring trinomials quadratic expressions (unless you are being taught things out of order in introductory algebra). Try that on each of the three equations, but you might find the last equation difficult to factor.

for graphing, you can simply create tables for points (x, X) (why are you using capital X on the left side?); and from the table, plot points on cartesian system. Finding the zeros might be difficult this way, but if you have the factorization, then set 0=binomial*binomial, and find the x value fow which each binomial is 0.

For quadratic equation, Feldoh gave you the right method, so just reread that.